3 Answers
3

The area of $\Delta OAC$ is $\frac 12 \cdot 1 \cdot \sin x=\frac 12\sin x$. The area of the circular sector $AOC$ is $\frac x2$. The area of triangle $\Delta OAB$ is $\frac12\cdot 1\cdot \tan x=\frac 12\tan x$. Noting that these shapes are contained within each other (and rewriting the variable $x$ as $\theta$ for consistency), we write

Take $\lim_{\theta\to 0^+}\cos\theta$ and apply the squeeze theorem. Just flip the inequalities around in the first line for the negative case. When you go to the second line by dividing by $\sin\theta$, which is negative then $\theta$ is negative, the rest of the argument proceeds the exact same way.

@JohnJoy what's your definition of $\pi$ to start with? There are plenty that don't need to make any reference to trig functions whatsoever. They might assume facts that end up being logically equivalent to $\lim_{x\to 0} \sin x/x=1$, but I don't see how it's inevitably circular.
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Robert MastragostinoJul 1 '14 at 22:09

I've always disliked this justification, because it is almost always the case that one shows that the derivative of $\sin$ is $\cos$ using the limit asked about, meaning that this is (likely) circular.
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mixedmath♦Aug 31 '12 at 3:45

if you dislike this, try power series version : $sin(h)=h-h^3/3!+h^5/5!+...$, then $sin(h)/h=1-h^2/3!+h^4/5!+...$, and the limit when h goes to zero then is 1.
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saraAug 31 '12 at 5:07

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@Solmaz: Again this is circular. The Taylor series of $sin x$ requires one to know derivatives of $\sin x$.
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JavaManAug 31 '12 at 7:10

1

@JavaMan: I believe that sara is using the power series as the definition for $\sin$ and $\cos$.
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robjohn♦Mar 27 '13 at 17:45